Why is "pulling the constant out of the limit" a valid algebra step in deriving the infinite geometric sum?

It's legal because of a fundamental property of limits: the limit of a product equals the product of the limits, provided each limit exists.

The formal rule:

If $g(n)$ is a constant (doesn't depend on $n$), then $\lim g(n) = g$, and the rule becomes:

Applied to the geometric sum:

Here $a/(1 - r)$ is a constant (doesn't involve $n$). So:

If $|r| < 1$, then $\lim r^n = 0$, so $\lim (1 - r^n) = 1$.

Therefore: $S_\infty = \dfrac{a}{1 - r} \cdot 1 = \dfrac{a}{1 - r}$.

Why this is non-trivial:

The rule fails if either limit doesn't exist. For example, if $f(n) = (-1)^n$ (oscillates between $\pm 1$) and $g(n)$ = a constant, then $\lim_{n \to \infty} [g \cdot f(n)]$ does NOT equal $g \cdot \lim f(n)$ because $\lim f(n)$ doesn't exist.

In the geometric series case, $\lim (1 - r^n)$ exists (it equals 1 when $|r| < 1$, or doesn't exist when $|r| \ge 1$). So we can only pull the constant out when convergence holds.

The "pull the constant out" intuition:

A constant is just a number that doesn't change. When you compute a limit, you're asking "where does the expression go as $n$ grows?" A constant doesn't go anywhere — it stays put. So multiplying by a constant just rescales whatever the limit is.

For an analogy: if your bank balance grows by $5\%$ per year, the growth rate is independent of the initial deposit. Pulling out the initial deposit (the constant) doesn't change the growth pattern — it just rescales the final number.

Why students get confused:

The notation $S_n = a(1 - r^n)/(1 - r)$ hides the structure. Once you rewrite as $S_n = [a/(1-r)] \times (1 - r^n)$, it becomes obvious that the first factor is invariant under the limit operation.

This is the same pattern as taking $E[c \cdot X] = c \cdot E[X]$ in probability, or $\int c \cdot f(x)\,dx = c \cdot \int f(x)\,dx$ in calculus. Pulling constants out is universal in linear operators.

For the exam:

You don't need to derive $S_\infty$ from first principles. Memorise $S_\infty = a/(1-r)$, know it requires $|r| < 1$, and recognise the algebra step when you see it. But understanding the derivation makes you more confident handling unfamiliar formulations.